{"id":6069,"date":"2023-09-08T04:35:50","date_gmt":"2023-09-08T04:35:50","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6069"},"modified":"2023-09-08T04:35:50","modified_gmt":"2023-09-08T04:35:50","slug":"ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-1\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1"},"content":{"rendered":"<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><br \/>\n<strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 8 (Introduction to Trigonometry)\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 8 Introduction to Trigonometry\u00a0 <\/strong>Exercise 8.1 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter : 8 Introduction to Trigonometry<\/strong> <\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 8.2<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-3\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 8.3<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-4\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 8.4<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 8.1<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. In \u2206 ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:<br \/>\n<\/strong>(i) sin A, cos A<br \/>\n(ii) sin C, cos C<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong> \u00a0In a given triangle ABC, right angled at B = \u2220B = 90\u00b0<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6480\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex8.1-Ans1i.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"330\" height=\"256\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex8.1-Ans1i.png 330w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex8.1-Ans1i-300x233.png 300w\" sizes=\"auto, (max-width: 330px) 100vw, 330px\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Given :<\/strong> AB = 24 cm and BC = 7 cm<br \/>\nAccording to the Pythagoras Theorem,<br \/>\nAC<sup>2<\/sup> = AB<sup>2<\/sup>\u00a0+ BC<sup>2<br \/>\n<\/sup>= 24<sup>2<\/sup>\u00a0+ 7<sup>2<\/sup><br \/>\n= 576 + 49 cm<sup>2<\/sup><br \/>\n= 625 cm<sup>2<\/sup><br \/>\nAC = 25<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i)<\/strong> <strong>sin A =<\/strong> BC\/AC = 7\/25<br \/>\n<strong>cos A =<\/strong> AB\/AC = 24\/25<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) sin C<\/strong> = AB\/AC = 24\/25<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>cos C<\/strong> = BC\/AC = 7\/25<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>2. In Fig. 8.13, find tan P \u2013 cot R<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6481\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex8.1-Que2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"185\" height=\"249\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong> \u00a0In the given \u0394PQR, the given triangle is right angled at Q and the given measures are:<br \/>\nPR = 13cm,<br \/>\nPQ = 12cm<br \/>\nAccording to Pythagorean theorem,<br \/>\nPR<sup>2<\/sup> = PQ<sup>2<\/sup> + QR<sup>2<br \/>\n<\/sup>(13 cm)<sup>2<\/sup> = (12 cm)<sup>2<\/sup> + QR<sup>2<br \/>\n<\/sup>169 cm<sup>2<\/sup> = 144 cm<sup>2<\/sup> + QR<sup>2<br \/>\n<\/sup>25 cm<sup>2<\/sup> = QR<sup>2<br \/>\n<\/sup>QR = 5 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">tan P = QR\/PQ<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">tan P = 5\/12<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">cot R = QR\/PQ<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">cot R = 5\/12 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">tan P \u2013 cot R = 5\/12 \u2013 5\/12 = 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3. If sin A = 3\/4, calculate cos A and tan A.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong> \u00a0Let us assume a right angled triangle ABC, right angled at B<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6482\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex8.1-Ans3.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"193\" height=\"241\" \/><br \/>\n<strong>Given :<\/strong> Sin A = 3\/4<br \/>\nBC\/AC = 3\/4<br \/>\nTherefore, hypotenuse AC will be 4k where k is a positive integer.<br \/>\nApplying Pythagoras theorem on \u2206ABC, we obtain:<br \/>\nAC<sup>2<\/sup>\u00a0= AB<sup>2<\/sup>\u00a0+ BC<sup>2<br \/>\n<\/sup>AB<sup>2<\/sup> = AC<sup>2<\/sup> &#8211; BC<sup>2<br \/>\n<\/sup>AB<sup>2\u00a0<\/sup>= (4k)<sup>2<\/sup> &#8211; (3k)<sup>2<br \/>\n<\/sup>AB<sup>2<\/sup> = 16k<sup>2<\/sup> &#8211; 9k<sup>2<br \/>\n<\/sup>AB<sup>2<\/sup> = 7 k<sup>2<br \/>\n<\/sup>AB = \u221a7 k<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Now, we have to find the value of cos A and tan A<br \/>\ncos A = AB\/AC\u00a0 = \u221a7k\/4k = \u221a7\/4<br \/>\ntan A = BC\/AB = 3k\/\u221a7k = 3\/\u221a7<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4. Given 15 cot A = 8, find sin A and sec A.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong> \u00a0Let us assume a right angled triangle ABC, right angled at B<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6483\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex8.1-Ans4.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"192\" height=\"231\" \/><br \/>\n<strong>Given :<\/strong> 15 cot A = 8<br \/>\nCot A = 8\/15<br \/>\nAB\/BC = 8\/15<br \/>\nApplying Pythagoras theorem in \u0394ABC, we obtain.<br \/>\nAC<sup>2<\/sup>\u00a0= AB<sup>2<\/sup> + BC<sup>2<br \/>\n<\/sup>AC<sup>2<\/sup> =(8k)<sup>2<\/sup> + (15k)<sup>2<br \/>\n<\/sup>AC<sup>2<\/sup> = 64k<sup>2<\/sup> + 225k<sup>2<br \/>\n<\/sup>AC<sup>2<\/sup> = 289k<sup>2<br \/>\n<\/sup>AC = 17k<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Now, we have to find the value of sin A and sec A<br \/>\nsin A = BC\/AC = 15k\/17k = 15\/17<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">sec A = AC\/AB = 17k \/8k = 17\/8<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>5. Given sec \u03b8 = 13\/12 Calculate all other trigonometric ratios <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0<\/strong>Let \u0394ABC be a right-angled triangle, right-angled at B.<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6484\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex8.1-Ans5.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"131\" height=\"188\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Given :\u00a0<\/strong> sec \u03b8 = 13\/12 = Hypotenuse\/Adjacent side = AC\/AB<br \/>\nLet AC be 13k and AB will be 12k<br \/>\nBy Pythagoras theorem we get,<br \/>\nAC<sup>2<\/sup>=AB<sup>2 <\/sup>+ BC<sup>2<br \/>\n<\/sup>(13k)<sup>2<\/sup>= (12k)<sup>2<\/sup> + BC<sup>2<br \/>\n<\/sup>169k<sup>2<\/sup>= 144k<sup>2<\/sup> + BC<sup>2<br \/>\n<\/sup>169k<sup>2<\/sup>= 144k<sup>2<\/sup> + BC<sup>2<br \/>\n<\/sup>BC<sup>2 = <\/sup>169k<sup>2<\/sup> \u2013 144k<sup>2<br \/>\n<\/sup>BC<sup>2<\/sup>= 25k<sup>2<br \/>\n<\/sup>Therefore, BC = 5k<br \/>\nNow, substitute the corresponding values in all other trigonometric ratios<br \/>\nsin \u03b8 = BC\/AC = 5\/13<br \/>\ncos \u03b8 = AB\/AC = 12\/13<br \/>\ntan \u03b8 = BC\/AB = 5\/12<br \/>\ncosec \u03b8 = AC\/BC = 13\/5<br \/>\ncot \u03b8 = AB\/BC = 12\/5<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>6. If \u2220A and \u2220B are acute angles such that cos A = cos B, then show that \u2220 A = \u2220 B. <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong> \u00a0Let us assume the triangle ABC in which CD \u22a5 AB<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6485\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex8.1-Ans6.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"144\" height=\"153\" \/><br \/>\nGive that the angles A and B are acute angles, such that<br \/>\ncos A = cos B<br \/>\nAs per the angles taken, the cos ratio is written as<br \/>\nAD\/AC = BD\/BC<br \/>\nAD\/BD = AC\/BC<br \/>\nLet take a constant value<br \/>\nAD\/BD = AC\/BC = k<br \/>\nNow consider the equation as<br \/>\nAD = k.BD &#8212;&#8212;&#8212;- (i)<br \/>\nAC = k BC &#8212;&#8212;&#8212;- (ii)<br \/>\nBy applying Pythagoras theorem in\u00a0\u25b3CAD\u00a0and\u00a0\u25b3CBD\u00a0we get,<br \/>\nCD<sup>2<\/sup> = BC<sup>2<\/sup> \u2013 BD<sup>2\u00a0\u00a0<\/sup>&#8212;&#8212;&#8212;- (iii)<br \/>\nCD<sup>2 <\/sup>=AC<sup>2 <\/sup>\u2212AD<sup>2<\/sup>\u00a0 \u00a0&#8212;&#8212;&#8212;- (iv)<br \/>\nFrom the equations (iii) and (iv) we get,<br \/>\nAC<sup>2<\/sup>\u2212AD<sup>2 <\/sup>= BC<sup>2<\/sup>\u2212BD<sup>2<br \/>\n<\/sup>\u21d2 (kBC)<sup>2<\/sup> &#8211; (kBD)<sup>2<\/sup>\u00a0= BC<sup>2<\/sup>\u00a0&#8211;\u00a0BD<sup>2<\/sup><br \/>\n\u21d2 k<sup>2<\/sup>(BC<sup>2<\/sup>\u00a0&#8211;\u00a0BD<sup>2<\/sup>) =\u00a0BC<sup>2<\/sup>\u00a0&#8211;\u00a0BD<sup>2<\/sup><br \/>\n\u21d2\u00a0k<sup>2<\/sup>\u00a0= 1<br \/>\n\u21d2 k = 1<br \/>\nPutting this value in equation (ii), we obtain<br \/>\nAC = BC<br \/>\n\u2220A = \u2220B (Angles opposite to equal sides of a triangle are equal-isosceles triangle)<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>7. If cot \u03b8 = 7\/8, evaluate :<br \/>\n<\/strong>(i) (1 + sin \u03b8)(1 \u2013 sin \u03b8)\/(1+cos \u03b8)(1-cos \u03b8)<br \/>\n(ii) cot<sup>2<\/sup> \u03b8<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong> \u00a0Let us assume a \u25b3ABC in which \u2220B = 90\u00b0 and \u2220C = \u03b8<br \/>\n<strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6484\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex8.1-Ans5.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"141\" height=\"203\" \/><br \/>\nGiven :<br \/>\n<\/strong>cot \u03b8 = BC\/AB = 7\/8<br \/>\nBC = 7k<br \/>\nAB = 8k<br \/>\nAccording to Pythagoras theorem in \u25b3ABC we get.<br \/>\nAC<sup>2 <\/sup>= AB<sup>2<\/sup>+BC<sup>2<br \/>\n<\/sup>AC<sup>2 <\/sup>= (8k)<sup>2 <\/sup>+ (7k)<sup>2<br \/>\n<\/sup>AC<sup>2 <\/sup>= 64k<sup>2 <\/sup>+ 49k<sup>2<br \/>\n<\/sup>AC<sup>2\u00a0 <\/sup>= 113k<sup>2<br \/>\n<\/sup>AC = \u221a113 k<br \/>\nAccording to the sin and cos function ratios, it is written as<br \/>\nsin \u03b8 = AB\/AC = 8k\/\u221a113k = 8\/\u221a113<br \/>\ncos \u03b8 = BC\/AC = 7k\/\u221a113 k = 7\/\u221a113<br \/>\nNow apply the values of sin function and cos function:<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\boldsymbol{\\frac{1+sin\\Theta&amp;space;(1-sin\\Theta)}{(1+cos\\Theta)(1-cos\\Theta)}}\" alt=\"\\boldsymbol{\\frac{1+sin\\Theta (1-sin\\Theta)}{(1+cos\\Theta)(1-cos\\Theta)}}\" align=\"absmiddle\" \/><br \/>\n<\/strong>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{(1-sin^2\\Theta)}{(1-cos^2\\Theta)}\" alt=\"\\frac{(1-sin^2\\Theta)}{(1-cos^2\\Theta)}\" align=\"absmiddle\" \/><br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1-\\left&amp;space;(&amp;space;\\frac{8}{\\sqrt{113}}&amp;space;\\right&amp;space;)^2}{1-\\left&amp;space;(&amp;space;\\frac{7}{\\sqrt{113}}&amp;space;\\right&amp;space;)^2}\" alt=\"\\frac{1-\\left ( \\frac{8}{\\sqrt{113}} \\right )^2}{1-\\left ( \\frac{7}{\\sqrt{113}} \\right )^2}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{113&amp;space;-64}{113-49}\" alt=\"\\frac{113 -64}{113-49}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{49}{64}\" alt=\"\\frac{49}{64}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) cot<sup>2<\/sup> \u03b8<\/strong><\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{7}{8}&amp;space;\\right&amp;space;)^2\" alt=\"\\left ( \\frac{7}{8} \\right )^2\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{49}{64}\" alt=\"\\frac{49}{64}\" align=\"absmiddle\" \/>\u00a0<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>8. If 3 cot A = 4, check whether (1-tan<sup>2 <\/sup>A)\/(1+tan<sup>2<\/sup> A) = cos<sup>2<\/sup> A \u2013 sin <sup>2 <\/sup>A or not.<\/strong><\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong> \u00a0Let\u00a0\u25b3ABC\u00a0in which\u00a0\u2220B=90\u00b0<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6486\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex8.1-Ans8.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"145\" height=\"181\" \/><br \/>\n3 cot A = 4<br \/>\ncot A = 4\/3 = AB\/BC<br \/>\nLet AB = 4k an BC =3k,<br \/>\nAccording to the Pythagorean theorem,<br \/>\nAC<sup>2 <\/sup>= AB<sup>2 <\/sup>+ BC<sup>2<br \/>\n<\/sup>AC<sup>2 <\/sup>= (4k)<sup>2 <\/sup>+ (3k)<sup>2<br \/>\n<\/sup>AC<sup>2 <\/sup>= 16k<sup>2 <\/sup>+ 9k<sup>2<br \/>\n<\/sup>AC<sup>2 <\/sup>= 25k<sup>2<br \/>\n<\/sup>AC = 5k<br \/>\nNow, apply the values corresponding to the ratios<br \/>\ntan A = BC\/AB = 3\/4<br \/>\nsin A = BC\/AC = 3\/5<br \/>\ncos A = AB\/AC = 4\/5<br \/>\nNow compare the left hand side(LHS) with right hand side(RHS)<br \/>\nL.H.S. = (1 &#8211; tan<sup>2<\/sup>A)\/(1 + tan<sup>2<\/sup>A)<br \/>\n= (1 &#8211; (3\/4)<sup>2<\/sup>)\/(1 + (3\/4)<sup>2<\/sup>)<br \/>\n= (1 &#8211; 9\/16)\/(1 + 9\/16)<br \/>\n= (16 &#8211; 9)\/(16 + 9)<br \/>\n= 7\/25<br \/>\nR.H.S. = cos<sup>2<\/sup>A\u00a0\u2013 sin<sup>2<\/sup>A<br \/>\n= (4\/5)<sup>2\u00a0<\/sup>&#8211;\u00a0(3\/4)<sup>2<\/sup><br \/>\n=<sup>\u00a0<\/sup>(16\/25)\u00a0&#8211;\u00a0(9\/25)<br \/>\n= 7\/25<br \/>\n<strong>R.H.S. = L.H.S.<\/strong><br \/>\nHence,<br \/>\n(1 &#8211; tan<sup>2<\/sup>A)\/(1 + tan<sup>2<\/sup>A) =\u00a0cos<sup>2<\/sup>A\u00a0\u2013 sin<sup>2<\/sup>A<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>9. In triangle ABC, right-angled at B, if tan A = 1\/\u221a3 find the value of:<br \/>\n<\/strong>(i) sin A cos C + cos A sin C<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">(ii) cos A cos C \u2013 sin A sin C<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong> \u00a0Let \u0394ABC in which\u00a0\u2220B=90\u00b0<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6484\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex8.1-Ans5.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"128\" height=\"184\" \/><br \/>\ntan A = BC\/AB = 1\/\u221a3<br \/>\nBC = 1k<br \/>\nAB = \u221a3 k,<br \/>\nBy Pythagoras theorem in \u0394ABC we get:<br \/>\nAC<sup>2 <\/sup>= AB<sup>2 <\/sup>+ BC<sup>2<br \/>\n<\/sup>AC<sup>2 <\/sup>= (\u221a3 k)<sup>2 <\/sup>+ (k)<sup>2<br \/>\n<\/sup>AC<sup>2 <\/sup>= 3k<sup>2 <\/sup>+ k<sup>2<br \/>\n<\/sup>AC<sup>2 <\/sup>= 4k<sup>2<br \/>\n<\/sup>AC = 2k<br \/>\nNow find the values of cos A, Sin A<br \/>\nSin A = BC\/AC = 1\/2<br \/>\nCos A = AB\/AC = \u221a3\/2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">Then find the values of cos C and sin C<br \/>\nSin C = AB\/AC = <strong>\u221a<\/strong>3\/2<br \/>\nCos C = BC\/AC = 1\/2<br \/>\nNow, substitute the values in the given problem<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) sin A cos C + cos A sin C<\/strong><br \/>\n= (1\/2) \u00d7 (1\/2 ) + \u221a3\/2 \u00d7 \u221a3\/2<br \/>\n= 1\/4 + 3\/4<br \/>\n= 1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) cos A cos C \u2013 sin A sin C<\/strong><br \/>\n= (<strong>\u221a<\/strong>3\/2 )(1\/2) \u2013 (1\/2) (<strong>\u221a<\/strong>3\/2 )<br \/>\n= 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>10. In \u2206 PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong> \u00a0In a given triangle PQR, right angled at Q, the following measures are<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6487\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex8.1-Ans10.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"177\" height=\"222\" \/><br \/>\nPQ = 5 cm<br \/>\nPR + QR = 25 cm<br \/>\nNow let us assume, QR = x<br \/>\nPR = 25 &#8211; QR<br \/>\nPR = 25- x<br \/>\nAccording to the Pythagorean Theorem,<br \/>\nPR<sup>2<\/sup> = PQ<sup>2<\/sup> + QR<sup>2<br \/>\n<\/sup>Substitute the value of PR as x<br \/>\n(25 &#8211; x)<sup> 2 <\/sup>= 5<sup>2 <\/sup>+ x<sup>2<br \/>\n<\/sup>25<sup>2<\/sup> + x<sup>2<\/sup> \u2013 50x = 25 + x<sup>2<br \/>\n<\/sup>625 + x<sup>2<\/sup>-50x -25 \u2013 x<sup>2 <\/sup>= 0<br \/>\n-50x = -600<br \/>\nx= -600\/-50<br \/>\nx = 12 = QR<br \/>\nNow, find the value of PR<br \/>\nPR = 25 &#8211; QR<br \/>\nPR = 25 &#8211; 12<br \/>\nPR = 13<br \/>\nNow, substitute the value to the given problem<br \/>\n<strong>sin p =<\/strong> QR\/PR = 12\/13<br \/>\n<strong>cos p<\/strong> = PQ\/PR = 5\/13<br \/>\n<strong>tan p<\/strong> = QR\/PQ = 12\/5<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>11. State whether the following are true or false. Justify your answer.<br \/>\n<\/strong>(i) The value of tan A is always less than 1.<br \/>\n(ii) sec A = 12\/5 for some value of angle A.<br \/>\n(iii) cos A is the abbreviation used for the cosecant of angle A.<br \/>\n(iv) cot A is the product of cot and A.<br \/>\n(v) sin \u03b8 = 4\/3 for some angle \u03b8.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong> <\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i)<\/strong> The value of tan A is always less than 1.\u00a0 <strong>(False)<\/strong><br \/>\nbecause tan 60<sup>\u00b0<\/sup> =\u00a0\u221a3\u00a0&gt; 1<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> sec A = 12\/5 for some value of angle A. <strong>(True)<br \/>\n<\/strong>As hypotenuse is the largest side, the ratio on RHS will be greater than 1. Hence, the value of sec A is always greater than or equal to 1. Thus, the given statement is true.<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii)<\/strong> cos A is the abbreviation used for the cosecant of angle A. <strong>(False)<\/strong><br \/>\nAbbreviation used for cosecant of \u2220A is cosec A and cos A is the abbreviation used for cosine of \u2220A. Hence the given statement is false.<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iv)<\/strong> cot A is the product of cot and A. <strong>(False)<\/strong><br \/>\ncot A is not the product of cot and A. It is the cotangent of \u2220A. Hence, the given statement is false.<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(v)<\/strong> sin \u03b8 = 4\/3 for some angle \u03b8. <strong>(False)<\/strong><br \/>\nWe know that in a right angled triangle, Hypotenuse is the longest side.<br \/>\n\u2234\u00a0sin \u03b8 will always less than 1 and it can never be\u00a04\/3 for any value of\u00a0\u03b8.<\/span><\/p>\n<p><span style=\"color: #0000ff; font-family: Georgia, Palatino;\"><b><i><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\">Go Back To Chapters<\/a><\/i><\/b><\/span><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 8 (Introduction to Trigonometry)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 8 Introduction to Trigonometry\u00a0 Exercise 8.1 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 8 Introduction to Trigonometry NCERT Class 10 Maths Solution [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1091,1092,1044,1049,1048],"class_list":["post-6069","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-8-introduction-to-trigonometry-solutions","tag-ncert-class-10-mathematics-exercise-8-1-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - 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